A hypothesis does not need to be correct for an experiment to be successful because
1 answer:
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<h3>Answer:</h3>
0.130 moles NH₃
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
[Given] 7.85 × 10²² molecules NH₃
[Solve] moles NH₃
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.130355 moles NH₃ ≈ 0.130 moles NH₃
Answer:
<u>∆H° reaction = -890.3 kJ</u>
Explanation:
The given equation is :
Now ,
O2 is in the standard state so its ∆H° is zero.
∆H° is calculated by considering the formation of CO2 , H2O and CH4 .
..........∆H°a = -393.5 kJ
.....∆H°b = -285.8 kJ
..........∆H°c = -74.8 kJ
Multiply equation of water H2O by 2
and reverse the direction of equation of CH4
Hence the sign of ∆H°c = +74.8 kJ becomes +ve.
We are doing this because CH4 is to be in the reactant side not in the product side.
∆H° reaction = ∆H°a +2(∆H°b) -∆H°c
∆H° reaction = -393.5 - 2(285.8) + 74.8
∆H° reaction = -890.3 kJ